Inclusion Mapping is Monomorphism
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Theorem
Let $\struct {S, \circ}$ be an algebraic structure.
Let $\struct {T, \circ}$ be an algebraic substructure of $S$.
Let $\iota: T \to S$ be the inclusion mapping from $T$ to $S$.
Then $\iota$ is a monomorphism.
Proof
We have that the inclusion mapping is an injection.
Now let $x, y \in T$:
\(\ds \map \iota {x \circ y}\) | \(=\) | \(\ds x \circ y\) | Definition of Inclusion Mapping | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \iota x \circ \map \iota y\) | Definition of Inclusion Mapping |
demonstrating that $\iota$ has the morphism property.
So $\iota$ is a homomorphism which is also an injection.
Thus by definition, $\iota$ is a monomorphism.
$\blacksquare$